2016 JGP Ljubljana Cup Ladies SP | Page 3 | Golden Skate

2016 JGP Ljubljana Cup Ladies SP

gsyzf

Medalist
Joined
Jan 15, 2015
I disagree.
JSF's strategy in this particular case is exactly opposite: based on their wish rather than the athletes' track record.
Track record, hard facts and statistics would suggest to reward skaters with better track record (Honda, Kihira) for their achievements in their first events by spreading them and giving BOTH OF THEM chance to win second event and qualify. Instead, it rewards skaters with worse records after first event with that chance, which is clearly unfair.

For example, I suspect it's theoretically possible, that before last event in Germany, Yuna Shiraiwa may have her JGPF spot entirely in her hands - in opposite to Marin and Rika before Lyublyana, one of whom will be at the mercy of other skaters' outcomes in last two events.
So JSF assignments strategy, can theoretically result in situation, where skater with 4th place after first event would be in better position to advance than skater with 2nd place after first event(!).

It just makes no sense mathematically, strategically and from fairness standpoint.

The silver medalist has a better chance to qualify than 4th place finishers regardless of which event she is assigned to. The silver medalists need at least 2nd to have a chance to qualify for JGPF. If the silver medalist wins the next event, she qualifies. If she finishes 2nd, she is at the mercy of the other gold & silver medalists' placements & scores in the last 3 events. The 4th place finisher need to win the next event, and even then, she cannot qualify unless all gold medalists placed lower than 3rd, all silver medalists placed lower than 2nd, all bronze medalists placed lower than 3rd and the other 4th place finishers place lower than 1st and score lower than her. She is at the mercy of pretty much everybody. The bronze medalist needs to win the next event and can only qualify when all gold medalists place lower than 2nd, all silver medalists place lower than 1st and all other bronze medalists place not win and score lower than her . The silver medalists still have more chance to qualify than bronze medalist and 4th place finishers. But the silver medalist who is assigned to an easier event has a better chance to qualify for JGPF than another silver medalist who is assigned to a tougher event.
 
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Layback11

Record Breaker
Joined
Nov 18, 2014
I'm really excited to see Eun Soo (my favorite Korean junior) and of course Marin. Also hoping Ashley Lin has a good debut, although I've basically given up on the American juniors...
 

topaz emerald

Match Penalty
Joined
Dec 1, 2015
At only 13 y.o. it will be a good test of how Lim handles her nerves, that's for sure! She'll have plenty of time to think about it in the locker room as well between her 6 minute warm-up and the time she performs (what a wonderful learning experience too). Having to get on the ice when Honda is exiting and then having to hear Honda's score may possibly rattle her but I for one am a big Lim fan and absolutely love how supportive and helpful Queen Yuna has been to the future skaters of Korea!

just watched her sp program....she reminds me of a shorter version of anna Pogorilaya, her style and all. https://www.youtube.com/watch?v=50jdJJwSnWw
 

topaz emerald

Match Penalty
Joined
Dec 1, 2015

petitebrie

On the Ice
Joined
Sep 10, 2015
She does have an undeniable pop flair to her :) May I ask....what does the song "let's have a kiki" mean? Is this korean pop music?

I have no idea what that music is about. But I am excellent Googler, and I found this https://www.youtube.com/watch?v=5_RV-x_4tVQ and this https://www.youtube.com/watch?v=eGCD4xb-Tr8

So according to the video, Kiki means party. It does sound like K-pop.

"Kiki" comes from American Black, Latino and LGBT slang. It's a kind of party or gathering for close friends where you can relax, gossip, laugh and vent your frustrations. :)
 

bosskil

Match Penalty
Joined
Aug 17, 2016
Come on Marin :cheer2:,
light up your face with gladness
Hide every trace of sadness
Although a tear may be ever so near
That's the time you must keep on trying
Smile, what's the use of crying?
You'll find that life is still worthwhile
If you just smile... :)
 

Crossover

All Hail the Queen
Record Breaker
Joined
Oct 31, 2014
Come on Marin :cheer2:,
light up your face with gladness
Hide every trace of sadness
Although a tear may be ever so near
That's the time you must keep on trying
Smile, what's the use of crying?
You'll find that life is still worthwhile
If you just smile... :)

:confused:

EDIT: Ah..I get it. Your post is about the lyrics of Marin's SP music "Smile". :)

https://www.youtube.com/watch?v=UN8oLGBNXpE
 
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moriel

Record Breaker
Joined
Mar 18, 2015
It's not assumption. It's math. If this scenario happens only by coincidence, i.e. assume it's entirely random, there are 6 skaters to be assigned to 3 events, then the probability of one particular skater being assigned to one particular event is 1/6=0.167. So the probability of the combination of 2 silver medalists assigned to the first event, 2 bronze medalists assigned to the 2nd event and 2 4th place finishers assigned to the 3rd event = (1/6)*(1/6)*(1/6)*(1/6)*(1/6)*(1/6)=0.0001286=0.01286%, which is close to 0.

Everything is possible in theory, but not everything is probable in practice and real life. It's ok for fans to make decisions and strategies based on their wish. But it's unwise and irrational for fed officials to develop their strategies only based on their wish rather than the athletes' track records. The track records are hard facts and statistics.

On probability

1. Supposing skaters are assigned at random, any combination of skaters if equally likely. There is no statistical difference between 2 silver medalists assigned to the first event, 2 bronze medalists assigned to the 2nd event and 2 4th place finishers assigned to the 3rd event and any other combination of assignments, Drawing conclusions from sheer probability of this event is meaningless - yeah, its unlikely, but so is any other outcome, and if any other combination of assignments happened, you could have said the same: "it is so unlikely and surely there is some design in that".
Imagine a die with 1000 sides. You roll it and get 1000 - thats unlikely, right? But, based on this, you cannot say the die is not fair.
You could think the dice is not fair if you kept rolling it and got 1000 5 out of 10 times, for example =)

2. On actual calcs. I'm not knowledgeable on how the skaters are assigned to JGP events, but suppose we have 6 different skaters (2 silver, 2 bronze and 2 4th places) and need to distribute them between 3 events. Also, suppose we can assign only 2 skaters to each event.
Then, the probability of one particular skater being assigned to one particular event is 1/6=0.167 - this is not right, because we have 2 spots at each event and the order doesnt matter. The correct probability is 1/3. Also, this only applies to the first skater assigned, as, once we assign 1 skater, we have less skaters to chose from, and less spots to assign them.

So, the probability of a silver medalist being assigned to 1st event: 2 / 6 (as we have 2 silver medalists and can assign any of them).
The prob. of assigning 2 silver medalists to 1st event: 2/6 * 1/5 (we assign one, and then the probability of assigning the other to same event is 1/5, as we now have 5 skaters to chose from).
Givent that we assigned 2 silver medalists to 1st, the prob. of assigning 2 bronze to the second event is 2/4 * 1/3 (we assign one of them, and then place the second one at the remaining spot).
And then 2 remaining skaters are automatically assigned to last event.
So we have: 2/6 * 1/5 * 2/4 * 1/3 = 1/90 = 0.0111 or 1.1%.
 

MalAssada

Medalist
Joined
Jun 28, 2014
On probability

1.
Imagine a die with 1000 sides. You roll it and get 1000 - thats unlikely, right? But, based on this, you cannot say the die is not fair.
You could think the dice is not fair if you kept rolling it and got 1000 5 out of 10 times, for example =)

Forgive me for asking this, considering my Maths are high school level and after graduating I took a year to let my brain rot. But each toss of a dice results in a different outcome not related to the last, so surely the chances of getting 1000 5 out of 10 times is the same of getting any other random numbers in the 10 tossings? It would still be fair, only a freaking coincidence.
 

CaroLiza_fan

EZETTIE LATUASV IVAKMHA
Record Breaker
Joined
Oct 25, 2012
Country
Northern-Ireland
Group 1 Warm-Up started:

https://www.youtube.com/watch?v=jhXmyNq-nys

Unfortunately, I will not be watching the whole competition. I am currently watching my fellow countryman Mark Allen playing snooker. But, because of tennis and cycling, the match has been relegated to the Eurosport Player. Which means it will not be archived. (Plus Eurosport Player is a temperamental thing...)

However, I am planning to dip in to watch Group 2. (I wonder why... :think: )

All the best to all the skaters taking part.

And good luck to those of you that are planning to watch the whole competition! 35 skaters :drama: Nearly 5 hours! :drama:

This is going to be a marathon!

CaroLiza_fan
 
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moriel

Record Breaker
Joined
Mar 18, 2015
Forgive me for asking this, considering my Maths are high school level and after graduating I took a year to let my brain rot. But each toss of a dice results in a different outcome not related to the last, so surely the chances of getting 1000 5 out of 10 times is the same of getting any other random numbers in the 10 tossings? It would still be fair, only a freaking coincidence.

Think of it as following: out of 1000 results, you got the same value 5 out of 10 times.
Lets look at 2 tosses first: the prob. of getting 1000 or any other number in 1 toss is 1/1000
Now, the probability of getting 2 equal values in 2 tosses is 1/1000 (because there are 1000 different possibilities of equal values, with probability 1/1000 * 1/1000 each)
and the probability of getting 2 different values is 999/1000

3 tosses:
P(getting 3 of the same) = 1/1000 * 1/1000 (two of the same, and then another one
P(3 different) = 999/1000 * 998 / 1000 (998 on second because we already used 2 values on first throws andnow want a different one)


So basically, the issue here is not getting a specific outcome but getting the same result many times: we got 1000 possibilities, its way more likely to get all values different, or maybe just one repetition - but 5 out of 10 becomes suspicious.
 

Interspectator

Record Breaker
Joined
Dec 25, 2012
Lozko's spins are soooo amazing. I like her black swan goth-doll look she has going. I'd get behind her 100% if her jumps weren't so spinny, but I love the way this program showcases her flexibility and grace.
 
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